C++11 转换可变参数模板以推导 class 类型

Question

我想从 class 属性列表中生成 classes。对于每一个属性，实现都是通过一个模板推导出来的。

为了这个问题，我将通过尝试创建一个 class 来处理具有两个属性的虚拟实体来说明我的问题：治疗师和商人

#include <iostream>

// Base class for entities
class entity
{
    public:
    virtual void do_walk() { std::cout << "I like walking" << std::endl; };
};

// A class property for healer characters
class healer
{
    public:
    // interface
    virtual void do_heal() = 0;
};

// A class property for healer npcs
class merchant
{
    public:
    // interface
    virtual void do_sell() = 0;
};

// implementation of the healer property
class healer_implementation : public healer
{
    public:
    virtual void do_heal() { std::cout << "I heal people" << std::endl; }
};



// implementation of the healer property
class merchant_implementation : public merchant
{
    public:
    virtual void do_sell() { std::cout << "I sell things" << std::endl; }
};

// To deduce the implementation of a property, we'll use the template property which will be specialized for each property and reference an implementation
template<typename T>
class property
{};

template<>
class property<merchant>
{
public: typedef merchant_implementation implementation;
};

template<>
class property<healer>
{
public: typedef healer_implementation implementation;
};


// This class is a class helper to deduce the right class type from a list of property for an entity
template<typename PROPERTY0, typename PROPERTY1=void>
class factory
{
public:
    typedef typename property<PROPERTY0>::implementation base0;
    typedef typename property<PROPERTY1>::implementation base1;


    class custom : public base0, public base1, public entity {};

};

int main()
{
    entity* bob = new factory<healer, merchant>::custom();

    // bob will try to sell things
    merchant* m = dynamic_cast<merchant*>(bob);
    if (m)
        m->do_sell();

    // bob will try to heal people
    healer* h = dynamic_cast<healer*>(bob);
    if (h)
        h->do_heal();

    // and as an entity, bob can walk
    bob->do_walk();

return 1;
}

如果我执行这段代码，我得到了预期的结果：

I sell things
I heal people
I like walking

现在，如果我想创建一个只有一个属性的实体，我必须为工厂添加一个专业化：

template<typename PROPERTY0>
class factory<PROPERTY0, void>
{
public:
    typedef typename property<PROPERTY0>::implementation base0;

    class custom : public base0, public entity {};

};

我可以这样测试：

entity* joe = new factory<merchant>::custom();
h = dynamic_cast<healer*>(joe);
if (!h)
   std::cout << "Joe is not an healer." << std::endl;


m = dynamic_cast<merchant*>(joe);
if (m)
    m->do_sell();

然后我得到：

Joe is not an healer
I sell things

我的问题是：有没有一种方法可以使用可变参数模板对 class 工厂进行编码以处理任意数量的属性，或者如果我需要一个最多具有N 个属性 ?

Answer 1

如果你可以不用 typedefs base0、base1 等，你确实可以做到这一点：

template <class... Property>
class factory
{
public:
    typedef typename property<PROPERTY0>::implementation base0;
    typedef typename property<PROPERTY1>::implementation base1;


    class custom : public property<Property>::implementation..., public entity {};

};

[Live example]

C++11 转换可变参数模板以推导 class 类型

C++11 Translating a variadic template to deduce a class type

variadic-templates

c++11